Exercise 5.9 An NPN amplifier driven from a voltage source

a)

b)

An answer:
For this specific circuit, a non-Sudoku-like derivation is shown below. The base current CANNOT be neglected: that is explicitly stated. $$\begin{array}{rcll}{R}_1& =& \frac{V_B}{I_{R1}}& \\ & & V_B=0.6+R_{3b}\cdot I_C(1+1∕\mathit{\beta fe})& \\ & & I_{R1}=I_2-I_C∕\mathit{\beta fe}& \\ R_1& =& \frac{0.6+R_{3b}\cdot I_C(1+1∕\mathit{\beta fe})}{I_2-I_C∕\mathit{\beta fe}}& \end{array}$$

c)

d)

e)